MHT CET 2024 15th May Evening Shift | Circle Question 3 | Mathematics

If the sides of a rectangle are given by the equations $x=-2, x=6, y=-2, y=5$, then the equation of the circle, drawn on the diagonal of this rectangle as its diameter, is

$x^2+y^2+4 x+3 y+22=0$

$x^2+y^2-4 x+3 y-22=0$

$x^2+y^2-4 x-3 y-22=0$

$x^2+y^2+4 x-3 y+22=0$

MHT CET 2024 15th May Morning Shift

MCQ (Single Correct Answer)

-0

The number of common tangents to the circles $x^2+y^2-x=0$ and $x^2+y^2+x=0$ is /are

MHT CET 2024 11th May Evening Shift

MCQ (Single Correct Answer)

-0

The parametric equations of the circle $x^2+y^2-\mathrm{a} x-b y=0$ are

$x=\frac{\mathrm{a}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{2} \cos \theta, y=\frac{\mathrm{b}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{2} \sin \theta$

$x=\frac{-\mathrm{a}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{4} \sin \theta, y=\frac{-\mathrm{b}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{4} \cos \theta$

$x=\frac{\mathrm{a}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{2}} \sin \theta, y=\frac{\mathrm{b}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{2}} \cos \theta$

$x=\frac{\mathrm{a}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{4} \cos \theta, y=\frac{\mathrm{b}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{4} \sin \theta$

MHT CET 2024 11th May Morning Shift

MCQ (Single Correct Answer)

-0

The equation of the tangent to the circle, given by $x=5 \cos \theta, y=5 \sin \theta$ at the point $\theta=\frac{\pi}{3}$ on it , is

$x-\sqrt{3} y=-5$

$x+\sqrt{3} y=10$

$\sqrt{3} x+y=5 \sqrt{3}$

$\sqrt{3} x-y=0$

PREVIOUS NEXT

MHT CET Subjects